What are the solutions to the equation 0= 1/3 t^2-6t+100 ?

The solutions to the equation 0= 1/3 t^2-6t+100 are t=9+sqrt(219)i,t=9-sqrt(219)i

Find the zeros of 0= 1/3 t^2-6t+100

The zeros of 0= 1/3 t^2-6t+100 are t=9+sqrt(219)i,t=9-sqrt(219)i

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0 = \frac{1}{3} t^{2} - 6 t + 100

t = 9 + 219 i, t = 9 - 219 i

Solution steps

0 = \frac{1}{3} t^{2} - 6 t + 100

Multiply both sides by

3

0 = t^{2} - 18 t + 300

Switch sides

t^{2} - 18 t + 300 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 300}{2 \cdot 1}

Simplify

(- 18)^{2} - 4 \cdot 1 \cdot 300 : 2219 i

t_{1, 2} = \frac{- ( - 18 ) \pm 2 219 i}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 18 ) + 2 219 i}{2 \cdot 1}, t_{2} = \frac{- ( - 18 ) - 2 219 i}{2 \cdot 1}

t = \frac{- ( - 18 ) + 2 219 i}{2 \cdot 1} : 9 + 219 i

t = \frac{- ( - 18 ) - 2 219 i}{2 \cdot 1} : 9 - 219 i

The solutions to the quadratic equation are:

t = 9 + 219 i, t = 9 - 219 i

rs + 4 s t

2 x ln (x) + x > 0

3 x (x - 2) > 1 - 7 x - 3 x^{2}

\frac{x + 2}{15} - \frac{x - 3}{5} = \frac{2}{3}

\frac{5}{( 7 - 2 x )} > 0

What are the solutions to the equation 0= 1/3 t^2-6t+100 ?
The solutions to the equation 0= 1/3 t^2-6t+100 are t=9+sqrt(219)i,t=9-sqrt(219)i
Find the zeros of 0= 1/3 t^2-6t+100
The zeros of 0= 1/3 t^2-6t+100 are t=9+sqrt(219)i,t=9-sqrt(219)i